Molar mass of LiBr (mm )= 86.845 g/mol
Molarity ( M ) = 4 M
Mass of solute ( m ) = 100 g
Volume ( V ) = in liters ?
V = m / mm * M
V = 100 / 86.845 * 4
V = 100 / 347.38
V = 0.2875 L
hope this helps!.
Answer:
A
Explanation:
Electron geometry describes the arrangement of electron groups. Molecular geometry describes the arrangement of atoms, excluding lone pairs.
Answer: The independent variable is the type of metal being used.
{Note: The "dependent variable" is the "measured density" that corresponds to each of the metals."}.
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Explanation:
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The "independent variable", which is plotted on the "x-axis" (horizontal axis), is the variable that can be "controlled/manipulated". In this case, this would be the type of metal chosen.
The "dependent variable" , which is plotted on the "y-axis" (vertical axis) is the "obtained value/measurement/result" (that "cannot be controlled/manipulated").
In this case, the "density", which is the "measured value" that corresponds to the selected "meal", is the "dependent variable".
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Hope this helpful to you!
Wishing you well!
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Answer:
Polar/Hydrophilic
Explanation:
Fluorine, Nitrogen and Oxygen are strong electronegative atoms and by definition, Electronegativity is the amount of pull or the high affinity of an atom to electrons.
Polar bond occurs when there is a high difference between the electronegativity value of both atoms that take part in the bond.
A polar molecule has a net dipole from the distribution of its positive and negayive charges. Hydrophobic and Hydrophilic (in chemistry, Polar) are terms dependent on the overall distribution of charge in its molecule.
Therefore, bonds between C-N, C-O and C-Cl are polar covalent bonds a d this is because of the jigh electronegativity possessed by Nitrogen, Oxygen and Chlorine.
The given concentration of boric acid = 0.0500 M
Required volume of the solution = 2 L
Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.
Calculating the moles of 0.0500 M boric acid present in 2 L solution:
Converting moles of boric acid to mass:
Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.
